Draw the triangles as shown in my first attached diagram. We draw the translation of the shape $|h|$ units to the right when $h>0$, and $|h|$ to the left when $h<0$, and $|k|$ units to the right when $k>0$, and $|k|$ units to the left when $k<0$. Using the distance formula, we find the distance between the points in the first triangle are $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}, \sqrt{(x_1-x_3)^2+(y_1-y_3)^2}, \sqrt{(x_2-x_3)^2+(y_2-y_3)^2}$. Using the distance formula on the second triangle, we find the distances are $\sqrt{\left( x_1+h-x_2-h\right)^2+(y_1+h-y_2-h)^2}, \sqrt{\left(x_2+h-x_3-h\right)^2+(y_2+h-y_3-h)^2}, \sqrt{(x_1+h-x_3-h)^2+(y_1+h-y_3-h)^2}= \sqrt{\left(x_1-x_2\right)^2+(y_1-y_2)^2}, \sqrt{(x_2-x_3)^2+(y_2-y_3)^2}, \sqrt{(x_1-x_3)^2+(y_1-y_3)^2}$, therefore the triangles are congruent by $SSS$. We apply shoelace theorem to find the area of the first triangle is $\frac{1}{2}|\left(x_1\times y_2+x_2\times y_3+x_3\times y_1\right)-\left(y_1\times x_2+y_2\times x_3+y_3\times x_1\right)|= $ \[\frac{1}{2}|x_1\times (y_2-y_3)+x_2\times (y_3-y_1)+x_3(y_1-y_2)|\] Now apply shoelace theorem to find the area of the second triangle is $\frac{1}{2}|\left((x_1+h)(y_2+k)+(x_2+h)(y_3+k)+(x_3+h)(y_1+k)\right)-$ $\left((y_1+k)(x_2+h)+(y_2+k)(x_3+h)+(y_3+k)(x_2+h)\right)|$ \[= \frac{1}{2}|\left((x_1+h)(y_2-y_3)+(x_2+h)(y_3-y_1)+(x_3+h)(y_1-y_2)\right)|\] Because the triangles are congruent, we know their areas are equal. Ignore the absolute value signs for now, and multiply both equations by $2$ to give: $\left((x_1+h)(y_2-y_3)+(x_2+h)(y_3-y_1)+(x_3+h)(y_1-y_2)\right)= x_1\times (y_2-y_3)+x_2\times (y_3-y_1)+x_3(y_1-y_2)$ Subtract $x_1\times (y_2-y_3)+x_2\times (y_3-y_1)+x_3(y_1-y_2)$ from both sides of the equation to result in (including factoring:) These equations are similar, with the only difference is the coefficients of the $y$ terms are $x_1+h$ instead of $x_1$. Because of this, we assume $h=0$. Assume that $x_1<x_2<x_3$ and $y_2<y_1$ and $y_3>y_2$( you can re-arrange the points in any such way, as long as the sub points* stay the same). During this case, we would have the vertical shift from $(x_2, y_2)$ to have to go over the top of the triangle. Because the $x$ coordinates are the same, the $y$ coordinate is going to have to go up over the top (or a distance of $y_3-y_2=\min(k)$ when $y_3>y_1$ and $y_1-y_2=\min(k)$ when $y_2>y_3$. Elsewhere, when the same restrictions apply on the $y$ coordinates, and $x_1<x_2<x_3$, we have no restrictions on $k$. We use these restrictions, and have the same assumptions on the boundaries of $x_1, x_2, x_3, y_1, y_2, y_3$. We know the area of the triangle is $\frac{1}{2}|x_1\times (y_2-y_3)+x_2\times (y_3-y_1)+x_3(y_1-y_2)|$ This proves to be quite useful with what we have discovered above. WLOG, $x_1< x_2< x_3, y_1>y_2$, and $y_3>y_2$. Because $y_3>y_2$, we know $(y_2-y_3)<0$, and because $y_1>y_2$ we have $y_1-y_2>0$. We do not know about the signs of $y_3$ and $y_1$, but again WLOG $y_3>y_1$. We put everything in the form $a(b-c)$ where $b>c$. $\frac{1}{2}|-x_1(y_3-y_2)+x_2\times (y_3-y_1)+x_3(y_1-y_2)|$ *By this I mean $x_2$ and $y_2$ would still have to correspond in the same order, and so forth.